$Phi(phi)=A e^{-im phi}+B e^{im phi}$ and $Theta(theta)=P_l^m(cos theta)$ are combined to $P_l^m(cos theta)...
$Phi(phi)=A e^{-im phi}+B e^{im phi}$ and $Theta(theta)=P_l^m(cos theta)$ are combined to form
$$P_l^m(cos theta) e^{im phi}$$
How?
Where did $A,B, e^{-im phi}$ go?
Does it read implicitly that one chooses $A,B$ s.t. they and $e^{-im phi}$ are eliminated?
http://mathworld.wolfram.com/SphericalHarmonic.html
spherical-harmonics
asked Dec 27 '18 at 13:34
mavavilj
2,68911032
add a comment |
$Phi(phi)=A e^{-im phi}+B e^{im phi}$ and $Theta(theta)=P_l^m(cos theta)$ are combined to form
$$P_l^m(cos theta) e^{im phi}$$
How?
Where did $A,B, e^{-im phi}$ go?
Does it read implicitly that one chooses $A,B$ s.t. they and $e^{-im phi}$ are eliminated?
http://mathworld.wolfram.com/SphericalHarmonic.html
spherical-harmonics
asked Dec 27 '18 at 13:34
mavavilj
2,68911032
add a comment |
$Phi(phi)=A e^{-im phi}+B e^{im phi}$ and $Theta(theta)=P_l^m(cos theta)$ are combined to form
$$P_l^m(cos theta) e^{im phi}$$
How?
Where did $A,B, e^{-im phi}$ go?
Does it read implicitly that one chooses $A,B$ s.t. they and $e^{-im phi}$ are eliminated?
http://mathworld.wolfram.com/SphericalHarmonic.html
spherical-harmonics
asked Dec 27 '18 at 13:34
mavavilj
2,68911032
$Phi(phi)=A e^{-im phi}+B e^{im phi}$ and $Theta(theta)=P_l^m(cos theta)$ are combined to form
$$P_l^m(cos theta) e^{im phi}$$
How?
Where did $A,B, e^{-im phi}$ go?
Does it read implicitly that one chooses $A,B$ s.t. they and $e^{-im phi}$ are eliminated?
http://mathworld.wolfram.com/SphericalHarmonic.html
spherical-harmonics
spherical-harmonics
asked Dec 27 '18 at 13:34
mavavilj
2,68911032
asked Dec 27 '18 at 13:34
mavavilj
2,68911032
asked Dec 27 '18 at 13:34
mavavilj
2,68911032
asked Dec 27 '18 at 13:34
mavavilj
2,68911032
asked Dec 27 '18 at 13:34
mavavilj
2,68911032
2,68911032
add a comment |
add a comment |
1 Answer
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The idea is to build a solution $u(theta,phi) = Theta(theta)Phi(phi)$ for the equation
$$
frac{Phi(phi)}{sin theta}frac{{rm d}}{{rm d}theta}left( sintheta frac{{rm d}Theta}{{rm d}theta}right) + frac{Theta(theta)}{sin^2theta} frac{{rm d}^2Phi}{{rm d}phi^2} + l(l + 1)Theta(theta)Phi(phi) = 0
$$
and you just found that $Phi(phi) = e^{imphi}$, $Theta(theta) = P^m_l(costheta)$ is one. That is
$$
u(theta, phi) = e^{imphi} P^m_l(costheta)
$$
is a solution. But so is $u^*(theta,phi)$. So in that sense they are combined
answered Dec 27 '18 at 14:23
caverac
13.8k21030
What you mean found out that $Phi(phi)=e^{im phi}$ is solution? The gen. sol is $Ae^{-im phi}+Be^{im phi}$.
– mavavilj
Dec 27 '18 at 15:11
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The idea is to build a solution $u(theta,phi) = Theta(theta)Phi(phi)$ for the equation
$$
frac{Phi(phi)}{sin theta}frac{{rm d}}{{rm d}theta}left( sintheta frac{{rm d}Theta}{{rm d}theta}right) + frac{Theta(theta)}{sin^2theta} frac{{rm d}^2Phi}{{rm d}phi^2} + l(l + 1)Theta(theta)Phi(phi) = 0
$$
and you just found that $Phi(phi) = e^{imphi}$, $Theta(theta) = P^m_l(costheta)$ is one. That is
$$
u(theta, phi) = e^{imphi} P^m_l(costheta)
$$
is a solution. But so is $u^*(theta,phi)$. So in that sense they are combined
answered Dec 27 '18 at 14:23
caverac
13.8k21030
What you mean found out that $Phi(phi)=e^{im phi}$ is solution? The gen. sol is $Ae^{-im phi}+Be^{im phi}$.
– mavavilj
Dec 27 '18 at 15:11
add a comment |
The idea is to build a solution $u(theta,phi) = Theta(theta)Phi(phi)$ for the equation
$$
frac{Phi(phi)}{sin theta}frac{{rm d}}{{rm d}theta}left( sintheta frac{{rm d}Theta}{{rm d}theta}right) + frac{Theta(theta)}{sin^2theta} frac{{rm d}^2Phi}{{rm d}phi^2} + l(l + 1)Theta(theta)Phi(phi) = 0
$$
and you just found that $Phi(phi) = e^{imphi}$, $Theta(theta) = P^m_l(costheta)$ is one. That is
$$
u(theta, phi) = e^{imphi} P^m_l(costheta)
$$
is a solution. But so is $u^*(theta,phi)$. So in that sense they are combined
answered Dec 27 '18 at 14:23
caverac
13.8k21030
What you mean found out that $Phi(phi)=e^{im phi}$ is solution? The gen. sol is $Ae^{-im phi}+Be^{im phi}$.
– mavavilj
Dec 27 '18 at 15:11
add a comment |
The idea is to build a solution $u(theta,phi) = Theta(theta)Phi(phi)$ for the equation
$$
frac{Phi(phi)}{sin theta}frac{{rm d}}{{rm d}theta}left( sintheta frac{{rm d}Theta}{{rm d}theta}right) + frac{Theta(theta)}{sin^2theta} frac{{rm d}^2Phi}{{rm d}phi^2} + l(l + 1)Theta(theta)Phi(phi) = 0
$$
and you just found that $Phi(phi) = e^{imphi}$, $Theta(theta) = P^m_l(costheta)$ is one. That is
$$
u(theta, phi) = e^{imphi} P^m_l(costheta)
$$
is a solution. But so is $u^*(theta,phi)$. So in that sense they are combined
answered Dec 27 '18 at 14:23
caverac
13.8k21030
The idea is to build a solution $u(theta,phi) = Theta(theta)Phi(phi)$ for the equation
$$
frac{Phi(phi)}{sin theta}frac{{rm d}}{{rm d}theta}left( sintheta frac{{rm d}Theta}{{rm d}theta}right) + frac{Theta(theta)}{sin^2theta} frac{{rm d}^2Phi}{{rm d}phi^2} + l(l + 1)Theta(theta)Phi(phi) = 0
$$
and you just found that $Phi(phi) = e^{imphi}$, $Theta(theta) = P^m_l(costheta)$ is one. That is
$$
u(theta, phi) = e^{imphi} P^m_l(costheta)
$$
is a solution. But so is $u^*(theta,phi)$. So in that sense they are combined
answered Dec 27 '18 at 14:23
caverac
13.8k21030
answered Dec 27 '18 at 14:23
caverac
13.8k21030
answered Dec 27 '18 at 14:23
caverac
13.8k21030
answered Dec 27 '18 at 14:23
caverac
13.8k21030
13.8k21030
What you mean found out that $Phi(phi)=e^{im phi}$ is solution? The gen. sol is $Ae^{-im phi}+Be^{im phi}$.
– mavavilj
Dec 27 '18 at 15:11
add a comment |
What you mean found out that $Phi(phi)=e^{im phi}$ is solution? The gen. sol is $Ae^{-im phi}+Be^{im phi}$.
– mavavilj
Dec 27 '18 at 15:11
What you mean found out that $Phi(phi)=e^{im phi}$ is solution? The gen. sol is $Ae^{-im phi}+Be^{im phi}$.
– mavavilj
Dec 27 '18 at 15:11
What you mean found out that $Phi(phi)=e^{im phi}$ is solution? The gen. sol is $Ae^{-im phi}+Be^{im phi}$.
– mavavilj
Dec 27 '18 at 15:11
add a comment |
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